8. n (n+1) (n+4).

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Solution

The n th term of the given series is,

a n =n( n+1 )( n+4 ) =n( n 2 +5n+4 ) = n 3 +5 n 2 +4n

The sum of the given series is,

S n = ∑ k=1 n a k = ∑ k=1 n ( k 3 +5 k 2 +4k ) = ∑ k=1 n k 3 + ∑ k=1 n 5 k 2 + ∑ k=1 n 4k = n 2 ( n+1 ) 2 4 + 5n( n+1 )( 2n+1 ) 6 + 4n( n+1 ) 2

Simplify further,

S n = n( n+1 ) 2 [ n( n+1 ) 2 + 5( 2n+1 ) 3 +4 ] = n( n+1 ) 2 [ 3 n 2 +3n+20n+10+24 6 ] = n( n+1 ) 2 [ 3 n 2 +23n+34 6 ] = n( n+1 )( 3 n 2 +23n+34 ) 12

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n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot 3 + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot 3^{2} - . . . . .,

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